Functional Continuous Uncertainty Principle

1. Introduction

For $h\in {\mathbb{C}}^d$, let ${\parallel h\parallel }_0$ be the number of nonzero entries in h. Let $\widehat{}:{\mathbb{C}}^d\to {\mathbb{C}}^d$ be the Fourier transform. Great and surprising inequality of Donoho and Stark which improved our life is the following.

Theorem 1.

(Donoho-Stark Uncertainty Principle) [1] For every $d\in \mathbb{N}$,

$${\left(\frac{{\parallel h\parallel }_0+{\parallel \widehat{h}\parallel }_0}{2}\right)}^2\ge {\parallel h\parallel }_0{\parallel \widehat{h}\parallel }_0\ge d,\forall h\in {\mathbb{C}}^d\setminus \left\{0\right\}.$$

(2)

In 2002, Elad and Bruckstein extended Inequality (2) to pairs of orthonormal bases [2]. Given a collection ${\left\{{\tau }_j\right\}}_{j=1}^n$ in a finite dimensional Hilbert space $\mathcal{H}$ over $\mathbb{K}$ ($\mathbb{R}$ or $\mathbb{C}$), we define

$${\theta }_{\tau }:\mathcal{H}\ni h\mapsto {\theta }_{\tau }h{\left(\langle h,{\tau }_j\rangle \right)}_{j=1}^n\in {\mathbb{K}}^n.$$

Theorem 2.

(Elad-Bruckstein Uncertainty Principle) [2] Let ${\left\{{\tau }_j\right\}}_{j=1}^n$, ${\left\{{\omega }_j\right\}}_{j=1}^n$ be two orthonormal bases for a finite dimensional Hilbert space $\mathcal{H}$. Then

$${\left(\frac{\parallel {\theta }_{\tau }{h\parallel }_0+{\parallel {\theta }_{\omega }h\parallel }_0}{2}\right)}^2\ge \parallel {\theta }_{\tau }{h\parallel }_0{\parallel {\theta }_{\omega }h\parallel }_0\ge \frac{1}{{\displaystyle \underset{1\le j,k\le n}{max}{\left|\langle {\tau }_j,{\omega }_k\rangle \right|}^2}},\forall h\in \mathcal{H}\setminus \left\{0\right\}.$$

In 2013, Ricaud and Torrésani showed that orthonormal bases in Theorem 2 can be improved to Parseval frames [3].

Theorem 3.

(Ricaud-Torrésani Uncertainty Principle) [3] Let ${\left\{{\tau }_j\right\}}_{j=1}^n$, ${\left\{{\omega }_j\right\}}_{j=1}^n$ be two Parseval frames for a finite dimensional Hilbert space $\mathcal{H}$. Then

$${\left(\frac{\parallel {\theta }_{\tau }{h\parallel }_0+{\parallel {\theta }_{\omega }h\parallel }_0}{2}\right)}^2\ge \parallel {\theta }_{\tau }{h\parallel }_0{\parallel {\theta }_{\omega }h\parallel }_0\ge \frac{1}{{\displaystyle \underset{1\le j,k\le n}{max}{\left|\langle {\tau }_j,{\omega }_k\rangle \right|}^2}},\forall h\in \mathcal{H}\setminus \left\{0\right\}.$$

In 2023, author made a major improvement of Theorem 3.

Theorem 4.

(Functional Donoho-Stark-Elad-Bruckstein-Ricaud-Torrésani Uncertainty Principle) [4] Let $({\left\{f_j\right\}}_{j=1}^n,{\left\{{\tau }_j\right\}}_{j=1}^n)$ and $({\left\{g_k\right\}}_{k=1}^m,{\left\{{\omega }_k\right\}}_{k=1}^m)$ be p-Schauder frames for a finite dimensional Banach space $\mathcal{X}$. Then for every $x\in \mathcal{X}\setminus \left\{0\right\}$, we have

$$\parallel {\theta }_f{x\parallel }_0^{\frac{1}{p}}\parallel {\theta }_g{x\parallel }_0^{\frac{1}{q}}\ge \frac{1}{{\displaystyle \underset{1\le j\le n,1\le k\le m}{max}\left|f_j\left({\omega }_k\right)\right|}}\mathrm{and}\parallel {\theta }_g{x\parallel }_0^{\frac{1}{p}}{\parallel {\theta }_{f}x\parallel }_0^{\frac{1}{q}}\ge \frac{1}{{\displaystyle \underset{1\le j\le n,1\le k\le m}{max}\left|g_k\left({\tau }_j\right)\right|}}.$$

(3)

By seeing paper [4], Prof. Philip B. Stark, Department of Statistics, University of California, Berkeley, asked the following question to the author.

Question 1.

(Philip B. Stark) What is the infinite dimensional version of Theorem [4]?

In this paper, we derive continuous uncertainty principle for Banach spaces which contains Theorem 4 as a particular case and also answers Question Section 1.

2. Functional Continuous Uncertainty Principle

In the paper, $\mathbb{K}$ denotes $\mathbb{C}$ or $\mathbb{R}$ and $\mathcal{X}$ denotes a Banach space (need not be finite dimensional) over $\mathbb{K}$. Dual of $\mathcal{X}$ is denoted by ${\mathcal{X}}^{*}$. Whenever $1<p<\infty$, q denotes conjugate index of p. We recall the notion of weak integral also known as Pettis integrals [5]. Let $(\Omega ,\mu )$ be a measure space and $\mathcal{X}$ be a Banach space. A function $f:\Omega \to \mathcal{X}$ is said to be weak integrable or Pettis integrable if following conditions hold.

 (i)

For every $\varphi \in {\mathcal{X}}^{*}$, the map $\varphi f:\Omega \to \mathbb{K}$ is measurable and $\varphi f\in {\mathcal{L}}^1(\Omega ,\mu )$.

(ii)

For every measurable subset $E\in \Omega$, there exists an (unique) element $x_E\in \mathcal{X}$ such that

$$\varphi \left(x_E\right)=\underset{E}{\int }\varphi \left(f\left(\alpha \right)\right)d\mu \left(\alpha \right),\forall \varphi \in {\mathcal{X}}^{*}.$$

The element $x_E\in \mathcal{X}$ is denoted by ${\int }_{E}f\left(\alpha \right)d\mu \left(\alpha \right)$. With this notion, we have

$$\varphi \left(\underset{E}{\int }f\left(\alpha \right)d\mu \left(\alpha \right)\right)=\underset{E}{\int }\varphi \left(f\left(\alpha \right)\right)d\mu \left(\alpha \right),\forall \varphi \in {\mathcal{X}}^{*},\forall E\in \Omega .$$

We need the following continuous version of p-Schauder frames [4].

Definition 1.

Let $(\Omega ,\mu )$ be a measure space. Let ${\left\{{\tau }_{\alpha }\right\}}_{\alpha \in \Omega }$ be a collection in a Banach space $\mathcal{X}$ and ${\left\{f_{\alpha }\right\}}_{\alpha \in \Omega }$ be a collection in ${\mathcal{X}}^{*}$. The pair $({\left\{f_{\alpha }\right\}}_{\alpha \in \Omega },{\left\{{\tau }_{\alpha }\right\}}_{\alpha \in \Omega })$ is said to be a continuous p-Schauder frame for $\mathcal{X}$ ($1<p<\infty$) if the following holds.

  (i) 

For every $x\in \mathcal{X}$, the map $\Omega \ni \alpha \mapsto f_{\alpha }\left(x\right)\in \mathbb{K}$ is measurable.

 (ii) 

For every $x\in \mathcal{X}$,

$${\parallel x\parallel }^p=\underset{\Omega }{\int }{\left|f_{\alpha }\left(x\right)\right|}^{p}d\mu \left(\alpha \right).$$

(iii) 

For every $x\in \mathcal{X}$, the map $\Omega \ni \alpha \mapsto f_{\alpha }\left(x\right){\tau }_{\alpha }\in \mathbb{K}$ is weakly measurable.

 (iv) 

For every $x\in \mathcal{X}$,

$$x=\underset{\Omega }{\int }{f}_{\alpha }\left(x\right){\tau }_{\alpha }d\mu \left(\alpha \right),$$

where the integral is weak integral.

We note that condition (i) in Definition 1 says that the map

$${\theta }_f:\mathcal{X}\ni x\mapsto {\theta }_{f}x\in {\mathcal{L}}^p(\Omega ,\mu );{\theta }_{f}x:\Omega \ni \alpha \mapsto \left({\theta }_{f}x\right)\left(\alpha \right)f_{\alpha }\left(x\right)\in \mathbb{K}$$

is a linear isometry. Following is the fundamental result of this paper.

Theorem 5.

(Functional Continuous Uncertainty Principle) Let $(\Omega ,\mu )$, $(\Delta ,\nu )$ be measure spaces. Let $({\left\{f_{\alpha }\right\}}_{\alpha \in \Omega },{\left\{{\tau }_{\alpha }\right\}}_{\alpha \in \Omega })$ and $({\left\{g_{\beta }\right\}}_{\beta \in \Delta },{\left\{{\omega }_{\beta }\right\}}_{\beta \in \Delta })$ be continuous p-Schauder frames for a Banach space $\mathcal{X}$. Then for every $x\in \mathcal{X}\setminus \left\{0\right\}$, we have

$$\begin{gathered} \mu {\left(\mathrm{supp}\left({\theta }_{f}x\right)\right)}^{\frac{1}{p}}\nu {\left(\mathrm{supp}\left({\theta }_{g}x\right)\right)}^{\frac{1}{q}}\ge \frac{1}{{\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|f_{\alpha }\left({\omega }_{\beta }\right)\right|}}, \\ \nu {\left(\mathrm{supp}\left({\theta }_{g}x\right)\right)}^{\frac{1}{p}}\mu {\left(\mathrm{supp}\left({\theta }_{f}x\right)\right)}^{\frac{1}{q}}\ge \frac{1}{{\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|g_{\beta }\left({\tau }_{\alpha }\right)\right|}}. \end{gathered}$$

(4)

Proof. 

Let $x\in \mathcal{X}\setminus \left\{0\right\}$ and q be the conjugate index of p. First using ${\theta }_f$ is an isometry and later using ${\theta }_g$ is an isometry, we get

$$\begin{array}{cc}\hfill {\parallel x\parallel }^p& =\parallel {\theta }_f{x\parallel }^p=\underset{\Omega }{\int }|f_{\alpha }{\left(x\right)|}^{p}d\mu \left(\alpha \right)=\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }{\left|f_{\alpha }\left(x\right)\right|}^{p}d\mu \left(\alpha \right)\hfill \\ \hfill & =\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }{\left|f_{\alpha }\left(\underset{\Delta }{\int }{g}_{\beta }\left(x\right){\omega }_{\beta }d\nu \left(\beta \right)\right)\right|}^{p}d\mu \left(\alpha \right)=\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }{\left|\underset{\Delta }{\int }{g}_{\beta }\left(x\right)f_{\alpha }\left({\omega }_{\beta }\right)d\nu \left(\beta \right)\right|}^{p}d\mu \left(\alpha \right)\hfill \\ \hfill & =\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }{\left|\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }{g}_{\beta }\left(x\right)f_{\alpha }\left({\omega }_{\beta }\right)d\nu \left(\beta \right)\right|}^{p}d\mu \left(\alpha \right)\le \underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }{\left(\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }\left|g_{\beta }\left(x\right)f_{\alpha }\left({\omega }_{\beta }\right)\right|d\nu \left(\beta \right)\right)}^{p}d\mu \left(\alpha \right)\hfill \\ \hfill & \le {\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|f_{\alpha }\left({\omega }_{\beta }\right)\right|}\right)}^p\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }{\left(\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }\left|g_{\beta }\left(x\right)\right|d\nu \left(\beta \right)\right)}^{p}d\mu \left(\alpha \right)\hfill \\ \hfill & ={\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|f_{\alpha }\left({\omega }_{\beta }\right)\right|}\right)}^p\mu \left(\mathrm{supp}\left({\theta }_{f}x\right)\right){\left(\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }\left|g_{\beta }\left(x\right)\right|d\nu \left(\beta \right)\right)}^p\hfill \\ \hfill & \le {\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|f_{\alpha }\left({\omega }_{\beta }\right)\right|}\right)}^p\mu \left(\mathrm{supp}\left({\theta }_{f}x\right)\right){\left(\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }{\left|g_{\beta }\left(x\right)\right|}^{p}d\nu \left(\beta \right)\right)}^{\frac{p}{p}}{\left(\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }{1}^{q}d\nu \left(\beta \right)\right)}^{\frac{p}{q}}\hfill \\ \hfill & ={\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|f_{\alpha }\left({\omega }_{\beta }\right)\right|}\right)}^p\mu \left(\mathrm{supp}\left({\theta }_{f}x\right)\right){\parallel {\theta }_{g}x\parallel }^p\nu {\left(\mathrm{supp}\left({\theta }_{g}x\right)\right)}^{\frac{p}{q}}\hfill \\ \hfill & ={\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|f_{\alpha }\left({\omega }_{\beta }\right)\right|}\right)}^p\mu \left(\mathrm{supp}\left({\theta }_{f}x\right)\right){\parallel x\parallel }^p\nu {\left(\mathrm{supp}\left({\theta }_{g}x\right)\right)}^{\frac{p}{q}}.\hfill \end{array}$$

Therefore

$$\frac{1}{{\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|f_{\alpha }\left({\omega }_{\beta }\right)\right|}}\le \mu {\left(\mathrm{supp}\left({\theta }_{f}x\right)\right)}^{\frac{1}{p}}\nu {\left(\mathrm{supp}\left({\theta }_{g}x\right)\right)}^{\frac{1}{q}}.$$

On the other way, first using ${\theta }_g$ is an isometry and ${\theta }_f$ is an isometry, we get

$$\begin{array}{cc}\hfill {\parallel x\parallel }^p& =\parallel {\theta }_g{x\parallel }^p=\underset{\Delta }{\int }|g_{\beta }{\left(x\right)|}^{p}d\nu \left(\beta \right)=\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }{\left|g_{\beta }\left(x\right)\right|}^{p}d\nu \left(\beta \right)\hfill \\ \hfill & =\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }{\left|g_{\beta }\left(\underset{\Omega }{\int }{f}_{\alpha }\left(x\right){\tau }_{\alpha }d\mu \left(\alpha \right)\right)\right|}^{p}d\nu \left(\beta \right)=\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }{\left|\underset{\Omega }{\int }{f}_{\alpha }\left(x\right)g_{\beta }\left({\tau }_{\alpha }\right)d\mu \left(\alpha \right)\right|}^{p}d\nu \left(\beta \right)\hfill \\ \hfill & =\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }{\left|\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }{f}_{\alpha }\left(x\right)g_{\beta }\left({\tau }_{\alpha }\right)d\mu \left(\alpha \right)\right|}^{p}d\nu \left(\beta \right)\le \underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }{\left(\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }\left|f_{\alpha }\left(x\right)g_{\beta }\left({\tau }_{\alpha }\right)\right|d\mu \left(\alpha \right)\right)}^{p}d\nu \left(\beta \right)\hfill \\ \hfill & \le {\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|g_{\beta }\left({\tau }_{\alpha }\right)\right|}\right)}^p\underset{\mathrm{supp}\left({\theta }_{g}x\right)}{\int }{\left(\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }\left|f_{\alpha }\left(x\right)\right|d\mu \left(\alpha \right)\right)}^{p}d\nu \left(\beta \right)\hfill \\ \hfill & ={\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|g_{\beta }\left({\tau }_{\alpha }\right)\right|}\right)}^p\nu \left(\mathrm{supp}\left({\theta }_{g}x\right)\right){\left(\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }\left|f_{\alpha }\left(x\right)\right|d\mu \left(\alpha \right)\right)}^p\hfill \\ \hfill & \le {\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|g_{\beta }\left({\tau }_{\alpha }\right)\right|}\right)}^p\nu \left(\mathrm{supp}\left({\theta }_{g}x\right)\right){\left(\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }{\left|f_{\alpha }\left(x\right)\right|}^{p}d\mu \left(\alpha \right)\right)}^{\frac{p}{p}}{\left(\underset{\mathrm{supp}\left({\theta }_{f}x\right)}{\int }{1}^{q}d\mu \left(\alpha \right)\right)}^{\frac{p}{q}}\hfill \\ \hfill & ={\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|g_{\beta }\left({\tau }_{\alpha }\right)\right|}\right)}^p\nu \left(\mathrm{supp}\left({\theta }_{g}x\right)\right){\parallel {\theta }_{f}x\parallel }^p\mu {\left(\mathrm{supp}\left({\theta }_{f}x\right)\right)}^{\frac{p}{q}}\hfill \\ \hfill & ={\left({\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|g_{\beta }\left({\tau }_{\alpha }\right)\right|}\right)}^p\nu \left(\mathrm{supp}\left({\theta }_{g}x\right)\right){\parallel x\parallel }^p\mu {\left(\mathrm{supp}\left({\theta }_{f}x\right)\right)}^{\frac{p}{q}}.\hfill \end{array}$$

Therefore

$$\frac{1}{{\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}\left|g_{\beta }\left({\tau }_{\alpha }\right)\right|}}\le \nu {\left(\mathrm{supp}\left({\theta }_{g}x\right)\right)}^{\frac{1}{p}}\mu {\left(\mathrm{supp}\left({\theta }_{f}x\right)\right)}^{\frac{1}{q}}.$$

□

Corollary 1.

Let $(\Omega ,\mu )$, $(\Delta ,\nu )$ be measure spaces. Let ${\left\{{\tau }_{\alpha }\right\}}_{\alpha \in \Omega }$ and ${\left\{{\omega }_{\beta }\right\}}_{\beta \in \Delta }$ be Parseval continuous frames for a Hilbert space $\mathcal{H}$. Then for every $h\in \mathcal{H}\setminus \left\{0\right\}$, we have

$$\mu \left(\mathrm{supp}\left({\theta }_{\tau }h\right)\right)\nu \left(\mathrm{supp}\left({\theta }_{\omega }h\right)\right)\ge \frac{1}{{\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}{\left|\langle {\omega }_{\beta },{\tau }_{\alpha }\rangle \right|}^2}},\nu \left(\mathrm{supp}\left({\theta }_{\omega }h\right)\right)\mu \left(\mathrm{supp}\left({\theta }_{\tau }h\right)\right)\ge \frac{1}{{\displaystyle \underset{\alpha \in \Omega ,\beta \in \Delta }{sup}{\left|\langle {\tau }_{\alpha },{\omega }_{\beta }\rangle \right|}^2}},$$

where

$$\begin{array}{cc}\hfill & {\theta }_{\tau }:\mathcal{H}\ni h\mapsto {\theta }_{\tau }h\in {\mathcal{L}}^2(\Omega ,\mu );{\theta }_{\tau }h:\Omega \ni \alpha \mapsto \left({\theta }_{\tau }h\right)\left(\alpha \right)\langle h,{\tau }_{\alpha }\rangle \in \mathbb{K},\hfill \\ \hfill & {\theta }_{\omega }:\mathcal{H}\ni h\mapsto {\theta }_{\omega }h\in {\mathcal{L}}^2(\Delta ,\nu );{\theta }_{\omega }h:\Delta \ni \beta \mapsto \left({\theta }_{\omega }h\right)\left(\beta \right)\langle h,{\omega }_{\beta }\rangle \in \mathbb{K}.\hfill \end{array}$$

Corollary 2.

Theorem 4 follows from Theorem 5.

Proof. 

Let $({\left\{f_j\right\}}_{j=1}^n,{\left\{{\tau }_j\right\}}_{j=1}^n)$ and $({\left\{g_k\right\}}_{k=1}^m,{\left\{{\omega }_k\right\}}_{k=1}^m)$ be p-Schauder frames for a finite dimensional Banach space $\mathcal{X}$. Define $\Omega \{1,\cdots ,n\}$ and $\Delta \{1,\cdots ,m\}$. Take $\mu$ as counting measure on $\Omega$ and $\nu$ as counting measure on $\Delta$. □

Theorem 5 brings the following question.

Question 2.

(i) Let $(\Omega ,\mu )$, $(\Delta ,\nu )$ be measure spaces, $\mathcal{X}$ be a Banach space and $p>1$. For which pairs of continuous p-Schauder frames $({\left\{f_{\alpha }\right\}}_{\alpha \in \Omega },{\left\{{\tau }_{\alpha }\right\}}_{\alpha \in \Omega })$ and  $({\left\{g_{\beta }\right\}}_{\beta \in \Delta },{\left\{{\omega }_{\beta }\right\}}_{\beta \in \Delta })$ for $\mathcal{X}$, we have equality in Inequality (4)?

(ii) 

What is the version of Theorem 5 for  $0<p\le 1$ and $p=\infty$?

If we can derive the uncertainty principles derived in [6] and [7] for p-Schauder frames, then we hope that we can get continuous versions of them.